Question

A seasoned mini golfer is trying to make par on a tricky number five hole. The golfer can complete the hole by hitting the ball from the flat section it lays on, up a 45∘ ramp, over a moat, and into the hole which is ????=1.45 m away from the end of the ramp. If the opening of the hole and the end of the ramp are at the same height, y=0.640 m, at what speed must the golfer hit the ball to launch the ball over the moat so that it lands directly in the hole? Assume a frictionless surface, so the ball slides without rotating. The acceleration due to gravity is 9.81 m/s2

Answer 1

Answer:5.17 m/s

Explanation:

Given

let u be the speed at cliff initial point

range over cliff is 1.45 m

and range of projectile is given by

$R=\frac{u^2\sin 2\theta }{g}$

$1.45=\frac{u^2\sin 90}{9.8}$

u=3.77 m/s

Conserving Energy

$E_{bottom}=E_{initial\ point\ at\ cliff}$

Kinetic energy=Kinetic energy +Potential energy gained

Let v be the initial velocity

$\frac{mv^2}{2}=mgh+\frac{mu^2}{2}$

$v^2=u^2+2gh$

$v=\sqrt{u^2+2gh}$

$v=\sqrt{3.77^2+2\time 9.8\times 0.64}$

$v=\sqrt{26.75}=5.17 m/s$

Answer 2

The speed at which the golfer must hit the ball to launch the ball over the moat so that it lands directly in the hole is;

v = 5.17 m/s

From law of conservation of energy, we can say that energy at bottom is equal to energy at initial point on the cliff.

Now, we are told the range of distance he hit the ball from the end of the ramp is 1.45. Thus; R = 1.45

We know that formula for range of projectile is;

R = (u² sin 2θ)/g

Thus, since we are told that θ = 45°, then;

1.45 = (u² sin (2 × 45))/9.81

1.45 = (u²/9.81)(sin 90)

sin 90 = 1

Thus;

u² = 1.45 × 9.81

u = √(1.45 × 9.81)

u = 3.77 m/s

Now, the total energy at initial point of cliff is;

Initial kinetic energy + initial potential energy.

Thus;

E_cliff = ½mu² + mgh

Also, total energy at bottom of cliff is;

E_bott = ½mv²

From earlier, using conservation of energy, we have;

½mv² = ½mu² + mgh

m will cancel out to give;

½v² = ½u² + gh

Multiply through by 2 to get;

v² = u² + 2gh

We are told that the opening of the hole and the end of the ramp are at the same height, y = 0.640

m. Thus; h = 0.64 m

Plugging in the relevant values;

v² = (1.45 × 9.8) + 2(9.81 × 0.64)

v² = 26.754

v = √26.754

v = 5.17 m/s

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